Question

An electron in the beam of a cathod-ray tube is accelerated by a potential difference of 2.12 kV . Then it passes through a region of transverse magnetic field, where it moves in a circular arc with a radius of 0.170 m .
What is the magnitude of the field?

Answer 1

Answer:

B=9.1397*10^-4 Tesla

Explanation:

To find the velocity first we put kinetic energy og electron is equal to potential energy of electron

K.E=P.E

$\frac{1}{2}*m*v^{2} =e*V$

where :

m is the mass of electron

v is the velocity

V is the potential difference

$v=\sqrt{\frac{2*e*V}{m} }$    eq 1

Radius of electron moving in magnetic field is given by:

$R=\frac{m*v}{q*B}$       eq 2

where:

m is the mass of electron

v is the velocity

q=e=charge of electron

B is the magnitude of magnetic field

Put v from eq 1 into eq 2

$R=\frac{m*\sqrt{\frac{2*e*V}{m} } }{e B}$

$B=\sqrt{\frac{2*m*V}{e*R^{2} } }$

$B=\sqrt{\frac{2*(9.31*10^{-31})*(2.12*10^{3}) }{(1.60*10^{-19})*(0.170)^{2} } }$

B=9.1397*10^-4 Tesla